Sulphur Cathode

Sulfur is a great cathode. problem isit is not electronically nor ionically conductive so it has to becombined with conductive additives and gelated by solvents in orderto function as a cathode (which reduces the active material loadingin the cathode). the theoretical capacity of sulfur is 1670 Ah/kg(compare to ~140 for lithium cobalt oxide cathode used in batteriesnow so 12 times higher). The "downside" is that a Li/Sbattery discharges at about 2V, or about 1/2 of your typical Li-ionbattery. So, lets assume you could obtain a more realistic Q=125Ah/kg capacity if you take into account the weight of othercomponents in the battery. If you discharge at 1C you get 125 A from1 kg battery. The power would be P=I x V = 125A x 2V = 250 W from 1kg sulfur. The energy would be E=Q x V = 125Ah x 2V = 250 Wh for 1 kgbattery. Now of course if you can keep the same capacity and higherdischarge rates (2C), then the power doubles to 500W.

This means that for 50 kWh of energyyou need 200 kg battery. So you can get the same driving range of theroadster with half the battery weight. You also get the same powerbased on V= I x R with only half the discharge rate 2C vs 4C forroadster which means higher safety.

Problems are that the voltages arelower (1/2) if you put the same number of cells in series as in theroadster battery pack, so the power and energy comes from a highercurrent being drawn which will be affected more by losses due tovarious resistances. In addition, the metal lithium anode is wellknow for dendritic formation so that will always be a stumblingpoint. However there is always hope for other anodes

Sulfur is a great cathode. problem is it is not electronically nor ionically conductive so it has to be combined with conductive additives and gelated by solvents in order to function as a cathode (which reduces the active material loading in the cathode). the theoretical capacity of sulfur is 1670 Ah/kg(compare to ~140 for lithium cobalt oxide cathode used in batteries now so 12 times higher). The "downside" is that a Li/S battery discharges at about 2V, or about 1/2 of your typical Li-ion battery. So, lets assume you could obtain a more realistic Q=125Ah/kg capacity if you take into account the weight of other components in the battery. If you discharge at 1C you get 125 A from1 kg battery. The power would be P=I x V = 125A x 2V = 250 W from 1kg sulfur. The energy would be E=Q x V = 125Ah x 2V = 250 Wh for 1 kg battery. Now of course if you can keep the same capacity and higher discharge rates (2C), then the power doubles to 500W.

This means that for 50 kWh of energy you need 200 kg battery. So you can get the same driving range of the roadster with half the battery weight. You also get the same power based on V= I x R with only half the discharge rate 2C vs 4C for roadster which means higher safety.

Problems are that the voltages are lower (1/2) if you put the same number of cells in series as in the roadster battery pack, so the power and energy comes from a higher current being drawn which will be affected more by losses due to various resistances. In addition, the metal lithium anode is well known for dendritic formation so that will always be a stumbling point. However there is always hope for other anodes

Yes, and half the cell voltage means twice the balancing efforts (and twice the reduction of function with an unbalanced pack).

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hi, i dont have an engineering background. could you please explain what a reduced voltage but a higher current mean for a high power battery pack like the roadsters? what dies the "balancing effort, etc" mean?

hi, i dont have an engineering background. could you please explain what a reduced voltage but a higher current mean for a high power battery pack like the roadsters? what dies the "balancing effort, etc" mean?

Sulfur is a great cathode. problem isit is not electronically nor ionically conductive so it has to becombined with conductive additives and gelated by solvents in orderto function as a cathode (which reduces the active material loadingin the cathode). the theoretical capacity of sulfur is 1670 Ah/kg(compare to ~140 for lithium cobalt oxide cathode used in batteriesnow so 12 times higher). The "downside" is that a Li/Sbattery discharges at about 2V, or about 1/2 of your typical Li-ionbattery. So, lets assume you could obtain a more realistic Q=125Ah/kg capacity if you take into account the weight of othercomponents in the battery. If you discharge at 1C you get 125 A from1 kg battery. The power would be P=I x V = 125A x 2V = 250 W from 1kg sulfur. The energy would be E=Q x V = 125Ah x 2V = 250 Wh for 1 kgbattery. Now of course if you can keep the same capacity and higherdischarge rates (2C), then the power doubles to 500W.

This means that for 50 kWh of energyyou need 200 kg battery. So you can get the same driving range of theroadster with half the battery weight. You also get the same powerbased on V= I x R with only half the discharge rate 2C vs 4C forroadster which means higher safety.

Problems are that the voltages arelower (1/2) if you put the same number of cells in series as in theroadster battery pack, so the power and energy comes from a highercurrent being drawn which will be affected more by losses due tovarious resistances. In addition, the metal lithium anode is wellknow for dendritic formation so that will always be a stumblingpoint. However there is always hope for other anodes

hi, i dont have an engineering background. could you please explain what a reduced voltage but a higher current mean for a high power battery pack like the roadsters? what dies the "balancing effort, etc" mean?

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Curious, here's your answers:

1). the original subject was about batteries that hold more energy per mass, and the specific issue was that the new batteries have a lower cell voltage. The idea is that for the same amount of power you'd need to double the number of cells in series - the reason is that power is a product of current and voltage. Therefore with half the cell voltage you'd need twice as many to reach the same pack voltage - I neglected to say that you could instead use a lower pack voltage and instead draw more current to achieve the same power -- but the whole reason that EVs use such high DC voltages is so that the current part is reduced; losses go as the square of the current, and only linearly with voltage.

Think of current as like a flow of water, and voltage as the height of pressure of the water. Pressure 'pushes' the water, but the flow actually makes things happen. This analogy is universally used in simple explanations but fails in the intricate details.

2). Balancing is an issue with any battery pack which consists of more than one cell in series. Remember that 'series' means that the voltage of the cells is additive (more emf 'pressure'), and this is what achieves a 300V pack voltage with 4V cells - there are 100 of them tied together, positive to negative as shown below. The problem with this configuration, even in something as simple as AA batteries in a gadget, is that slight variations between the cell construction, or charging history, can make the cell voltages vary from one another. This difference, if left to get bigger and bigger, will ultimately cause damage to one of the cells. Because of this, systems that do charge and discharge control monitor ALL the cells and if any one of them is too low (during discharge) or too high (during charge), the controller will stop the process. So, in my example, an unbalanced pack appears to become smaller and smaller in capacity due to the one (or more) cells that are too low or too high.

Charge balancing equipment evens out the charge between those cells, thus protecting the pack and maintaining a higher capacity. Cell balancing processes are interesting, but beyond the scope of this answer.

thank you for the kind explanation. in my opinion if possible 1 would try to obtain all the cells in series. currently this is not possible because the driving range and the energy would not be enough coming from 1 single cell. so in theory if you have 4 * the capacity could per cell, than you could have as an example 14 cells in parallel and 500 in series at 2 V each that is 1000 volts. in addition the 14 cells in parallel would provide the same capacity at the 69 sales currently in the Tesla battery pack. the total number of cells is the same, 7000. power is the same, energy is same, Max current is half so less wiring necessary. plus the battery is lighter because S is lighter, and orders of magnitude cheaper.

thank you for the kind explanation. in my opinion if possible 1 would try to obtain all the cells in series. currently this is not possible because the driving range and the energy would not be enough coming from 1 single cell. so in theory if you have 4 * the capacity could per cell, than you could have as an example 14 cells in parallel and 500 in series at 2 V each that is 1000 volts. in addition the 14 cells in parallel would provide the same capacity at the 69 sales currently in the Tesla battery pack. the total number of cells is the same, 7000. power is the same, energy is same, Max current is half so less wiring necessary. plus the battery is lighter because S is lighter, and orders of magnitude cheaper.

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There's a compromise to be made on the pack voltage. If the voltage is too high, you have problems with insulation and the voltage rating on the semiconductors used in the control gear (motor drive, charger etc.). If the voltage is too low, then the current becomes very large (as already discussed). So, for a particular model of cell the pack designer needs to decide how many to put in series and how many in parallel to get the necessary capacity and current.

If you suddenly had a new cell that was 'better' but had half the voltage yet double the current capability, then you would simply use the same number of cells and wire them with more in series and less in parallel.

Unfortunately, it's probably not as simple as that. The Model S pack is delicate balance of many aspects of the cells: energy density, power density, thermal characteristics, mechanical characteristics.

For example, if you had a new cell that was identical in all parameters to the current cells except double the Ah capacity (double energy density), then even that wouldn't be an ideal fit to the existing design. You could use it to make a 170kWh pack for the Model S, but you couldn't use it to make the 85kWh pack cheaper: using half the number of cells would give you half the maximum current and so limit the maximum performance (as with the existing 40kWh model). And you would still be limited to charging it at the current rate - so your 170kWh model would still only put on 150 miles in half an hour of supercharging (even though that's now only 25% rather than 50% of total capacity). Your cheaper 85kWh with half the cells would supercharge at half the rate of the current model. And the safety case might well be invalidated: the pack is designed to be safe with any one cell breaking down without setting fire to the whole pack: with double the energy in each cell, the calculations that make that work are probably invalid.

Any radical new chemistry is likely to be even less of a good fit - and so would need ground-up redesign, starting with the decision of what size cell to use as the basic unit and working up from there.

thank you for the reply. I guess my question is if I were to offer you a battery operated the 2 volts with af tcapacity of 150 to 200 Ah per kg. could u make a vehicle battery out of it? are there any fundamental red Flags with this battery for usage in high power applications is compared with current lithium
cobalt oxide batteries

thank you for the reply. I guess my question is if I were to offer you a battery operated the 2 volts with af tcapacity of 150 to 200 Ah per kg. could u make a vehicle battery out of it? are there any fundamental red Flags with this battery for usage in high power applications is compared with current lithium
cobalt oxide batteries

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2V/cell certainly isn't a red flag in itself - if all the other numbers are OK, then you just put twice as many in series (and correspondingly less in parallel) to get the voltage required.

But the other critical parameter is power density - the rate at which you can charge/discharge the cells.

So if we measure the energy density in Wh/Kg, and the power density in W/Kg, then for a given application these need to remain roughly in balance: if you are improving the Wh/Kg, then the W/Kg needs to remain at least the same (if you are using your new technology just to improve range), but preferably needs to improve by the same amount so that you can keep the range the same and reduce the weight/cost.

Currently, Tesla are using cells that favour energy density over power density - there are other cells they could have used (eg. those from A123 systems, I believe) that have higher W/Kg but worse Wh/Kg: if they had used those, then they would have needed more cells to reach the 85kWh target, and then had more power available than they need to run the motors even in the Performance model.

On the other hand, people building a series hybrid need less total energy (since they don't target the same range, having the ICE to recharge), but need about the same power to run their motors: they obviously want to use a smaller number/weight of cells, and so if they used the same cells as Tesla they wouldn't have enough power. Hence they use cells that favour power density over energy density.

I don't have any numbers for the cells Tesla are using, but A123's AHR32113 cell (advertised for hybrid vehicle applications) quotes energy density of 71Wh/Kg and power of 2700W/Kg.

The Model S has 85kWh and (performance model) needs 310kW at the motor, so if using those A123 cells it would need 1200Kg to meet the energy requirement, but only 115 Kg to meet the energy requirement. In fact, it uses cells with greater energy density and lower power density so that the two numbers meet in the middle.

When i envision energy density (E=V x Q, where Q is capacity) and power density (P=V x I) i immediately correlate the two by the discharge rate of the battery (C rate) for example if you discharge a material with an energy density of 50 Wh/kg at 1C (in 1 hour) it will deliver 50 W/kg. If the same material is discharged twice as fast at 2C, then it will deliver twice the power (100 W/kg). This can be understood if you consider the capacity is 25 Ah/kg (if it operates at 2V) so it can generate 25A if providing charge at a rate of 1C or 50A if providing charge at a rate of 2C.

a lot of battery chemistry is employed in enabling faster discharge kinetics (ions diffusing through the electrolyte, through various solid electrolyte interphases and through solid phases) while diminishing capacity fades or denaturation of electrodes.

Having said that, I would like to point out that a battery with a higher capacity, in addition to providing a high energy density due to its high Q term, will also be able to provide higher power at LOWER C rates, because more charges will be transported at the same time. Imagine a 100 Ah/kg material A and a 300 Ah/kg B material both being discharged at 1C. Material B will provide 3 times the charges even if discharged at the same C rate (1hour) and hence 3x the power.

in conclusion, an increase in capacity, especially in the case discussed here will also provide an advantage in power density, because lower C rates will provide the same power density. lower C rates usually result in higher lifetime (more charge/discharge cycles). In addition, post lithium ion batteries almost always benefit from higher capacities and a voltage reduction, hence my questions here for the esteemed engineering forum members.

Well, if you are keeping the C-rate constant while improving capacity then that's fine. Tesla are currently discharging at about 3.7C peak rate (but average in normal use much less than this - about 0.5C in fast highway driving). Supercharger max charge rate is about 1C.

So if you can keep those charge/discharge rates while boosting the energy density, then the electrical side is OK. 2V rather than 4V doesn't in itself matter for this application: it would be a disadvantage in much smaller applications (where you only have a single string of cells), but here there are so many cells that you can reorganize the parallel/series arrangement to get the desired pack voltage.

What's left is the mechanical side. If you have gone to 2V cells while maintaining the same energy density and C-rate, then the current per cell has already doubled. If you are then improving the energy density while keeping the C-rate, current per cell goes up even more. So the first question is whether you can get the necessary current out of the cell terminations, and then whether you can manage the cooling. Possibly the cell shape needs to change to optimise those aspects. There is also the safety issue: some of your energy density gains will be eaten up by extra metalwork in the pack structure (to provide enough passive cooling to ensure single cell failure does not spread to the whole pack): less Kg of actual cells, but increasing Kg of pack structure.

thank you arg. remember, C rate is up to the end user or designer engineer. the battery itself can achieve high rates without losing much capacity however. the whole advantage of capacity (without a major loss in voltage, ie 2V) is that you can use LOWER C rates to achieve the same power density. i appreciate the engineering point of view in this matter.